completion and heights of prime ideals

Question

Let $A$ be a noetherian, regular local domain of dimension $2$ (for instance the local ring at a smooth point of a surface) and consider its completion $\hat A$ at its maximal ideal. Now let's look at the canonical surjective map $$f:\rm{Spec}\,\hat A\rightarrow\rm{Spec}\, A$$

If $\mathfrak p$ is a prime ideal of $A$ of height $1$, can we conclude that the elements in $f^{-1}(\mathfrak p)$ have heights $1$?

If the answer is YES, I would like to know if this is a general rule. To be precise, if we drop the hypothesis "$A$ has dimension $2$" I would like to know if the height of $\mathfrak p$ is preserved in its fiber $f^{-1}(\mathfrak p)$.

Answer 1

The ring extension $A\subset\widehat A$ is flat hence it satisfies the going-down property and thus $\operatorname{ht}P\ge\operatorname{ht}\mathfrak p$ for any prime ideal $P\subset\widehat A$ lying over $\mathfrak p$. If $\operatorname{ht}P=2$ then $P$ equals the maximal ideal of $\widehat A$ (since $\dim\widehat A=2$) which lies over the maximal ideal of $A$, a contradiction.

This can be easily generalized to prime ideals of $A$ of height equal to $\dim A-1$.

In general, one can use the dimension formula and get $\dim A_{\mathfrak p}=\dim\widehat A_P+\dim\widehat A_P/\mathfrak p\widehat A_P$, so your question reduces to

Is $\dim\widehat A_P/\mathfrak p\widehat A_P=0$ ?

The answer is negative, in general. There are examples of regular local rings with $\dim\widehat A_P/\mathfrak p\widehat A_P>0$; see here.